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Full text of "Treatise On Analysis Vol-Ii"

11    THE SPACES L1 AND L2       159

hence also fg (with the above convention) (13.6.3), and so both sides of
(13,11.2.2) are zero. Next, the inequality is true when one of the factors on
the right-hand side is + oo (for the same reason). Hence we may assume that
both factors on the right-hand side of (13.11.2.2) are finite and nonzero,
which implies (13.6.4) that / and g are finite almost everywhere. Now if
a, b are any two real numbers ^ 0, we have

for all scalars / ^ 0. Bearing in mind (13.6.5) and (13.5.6), it follows that

2 fig dn  (\tf2 + r V) dfjL  t J*/2 dp + t~ 1 f V dp.
Putting t = N2(#)/N2(/), we have the inequality (13.11.2.2).

To prove (13.11.2.1) when p = 2, we may again restrict ourselves to
the case where N2(/) and N2(#) are finite, and / and g are ^ 0. Since
(/+ g)2 g 2(/2 + g2\ it follows that N2(/+ g) is finite. We may assume
that N2(/+ g) ^ 0. Moreover, we have

with the convention at the beginning of this section; hence ((13.5.6) and
(13.11.2.2))

(N2(/+ g)}2 = 1W+ g}2}  Ni(/(/+ #)) + N1(flr(/+ g))

 N2(/)N2(/ + g) + N2(<7)N2(/ + g)
from which (13.11.2.1) follows, by dividing through by N2(/+ g) * 0.

(13.11 .3)   If(fn) is any sequence of functions on X with values in R, then
(13.11.3.1)             N,   E/n   ^ f N,(/n)       (p = 1, 2).

n=l

This also has already been proved for/? = 1 (13.5.8). The proof is similar

(n        \ 2
]T fk \    is increasing and has
fc=i   /

as its upper envelope. Since (N^))1/2 g f N2(A), by (13.11.2.1)

and induction on n, we can apply (13.5.7) to the sequence (#) to obtain the
result.

We have already (13.7.3) defined the set JS?i(X, ju) of (finite) real-valued
ju-integrable functions, and we have seen that it is a vector space. Likewise,ous). With this